1) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow 4x^2 + 14x - 10x - 35=4x^2-25\)
\(\Leftrightarrow4x^2-4x^2+14x-10x=35-25\)
\(\Leftrightarrow4x=10\)
\(\Leftrightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}\)
2) \(x^2-4x+5\)
\(=-(4x-x^2-5 )\)
\(= -[-(x^2-4x)-5 ]\)
\(=-[ -(x^2-2x.2+4-4)-5 ]\)
\(= -[-(x-2)^2+4-5 ]\)
\(= -[-(x-2)^2-1 ]\)
Vì \(-(x-2)^2 ≤0\)\(\forall x\) \(\Rightarrow\) \(-(x-2)^2-1<0\) \(\forall x\)
\(\Rightarrow\)\(-[-(x-2)^2-1 ]>0\)\(\forall x\)
\(\Rightarrow x^2-4x+5>0\)\(\forall x\)
2
\(x^2-4x+5=x^2-4x+4+1\\ =\left(x-2\right)^2+1>0\)
1
\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\\ 10x-4x^2-14x+35=4x^2-25\\ 8x^2+4x-60=0\\ x^2+\dfrac{1}{2}x-\dfrac{60}{8}=0\\ x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}=\dfrac{1}{16}+\dfrac{60}{8}\\ \left(x+\dfrac{1}{4}\right)^2=\dfrac{121}{16}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{4}=\dfrac{11}{4}\\x+\dfrac{1}{4}=-\dfrac{11}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{4}\\x=-\dfrac{12}{4}=-3\end{matrix}\right.\)
1) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Rightarrow\)\(-4x-4x^2+35=4x^2-25\)
\(\Rightarrow\) \(-4x=4x^2-25-35+4x^2\)
\(\Rightarrow\)\(-4x=8x^2-60\)
\(\Rightarrow\)-4x - 8x2 = -60
\(\Rightarrow\)- 2x . ( 2 + 4x) = -60
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